For $n\in N,n\geq 2$, prove that $\displaystyle \sum_{k=1}^{n}(\dfrac{1}{2k-1}-\dfrac{1}{2k})>\dfrac {2n}{3n+1}$.

For $n\in N,n\geq 2$, prove that
$\displaystyle \sum_{k=1}^{n}(\dfrac{1}{2k-1}-\dfrac{1}{2k})>\dfrac {2n}{3n+1}$.

Note that

$\displaystyle 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\cdots+\frac{1}{2n-1}-\frac{1}{2n}$

$\displaystyle =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}$

$\displaystyle =\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right)$

$\displaystyle =\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right)$

$\displaystyle =\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)$

$\displaystyle \small =\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)$

$\displaystyle =\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$

Therefore

$\displaystyle \begin{align*}\sum_{k=1}^{n}\left(\dfrac{1}{2n-1}-\dfrac{1}{2n}\right)&=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\\&\ge \frac{(\overbrace{1+1+\cdots+1}^{\text{n times}})^2}{n+1+n+2+\cdots+2n}\,\,\small\text{(by the extended Cauchy-Schwarz inequality)}\\&=\frac{n^2}{\frac{n}{2}\left(n+1+2n\right)}\\&=\frac{2n}{3n+1}\,\,\,\,\text{(Q.E.D.)}\end{align*}$